Why is $(p’-p)^2 = -|mathbf {p’} - mathbf p|^2 + mathcal O(mathbf p^4)$?

The confusion comes because they took the low momentum approximation early. The four-vector should look like, $p = (E,textbf{p}) = (sqrt{m^2 + textbf{p}^2},textbf{p})$

When the momentum is low you can approximate the energy as $E approx m$, which is why they write their 4-vector as $p=(m,textbf{p})$. Using the full four-vector you should get,

begin{align} (p'-p)^2 &= p'^2 + p^2 -2p_mu p^mu &= m^2 + m^2 - 2(E'E - textbf{p}'cdottextbf{p}) &=2m^2 -2sqrt{(m^2+textbf{p}'^2)(m^2+textbf{p}^2)} + 2textbf{p}'cdottextbf{p} &=2m^2 -2m^2sqrt{1+frac{textbf{p}'}{m^2}^2+frac{textbf{p}^2}{m^2}+frac{textbf{p}'^2textbf{p}^2}{m^4}} + 2textbf{p}'cdottextbf{p} end{align} Then if $textbf{p}^2ll m^2$ and $textbf{p}'^2 ll m^2$ then we can simplify, begin{align} (p'-p)^2 &approx 2m^2 - 2m^2 -2m^2bigg(frac{textbf{p}^2}{2m^2}+frac{textbf{p}'^2}{2m^2}bigg) +mathcal O(textbf{p}^4) + 2textbf{p}'cdot textbf{p} &approx -textbf{p}'^2 +2textbf{p}'cdottextbf{p}-textbf{p}^2 = -|textbf{p}'-textbf{p}|^2 end{align} I was admittedly a little sloppy with my notation for the Taylor expansion in treating $textbf{p}$ and $textbf{p}'$ like the same variable.